When I solve the differential equation
$$y''(x)-y(x)=0$$
using the auxiliary equation, I get $$y(x)=Ae^{x}+Be^{-x}.$$ Is $$y(x)=A\sinh(x)+B\cosh(x)$$ just as good of a solution? How about $$y(x)=A\cosh(x)+B\cosh(x),$$ etc.?
Does it have infinitely many general solutions? Is this normal?
Thanks. :)
On
Your auxiliary equation is correct. Then we can write:
$$y=Ae^x+Be^{-x} $$
$$A'\cosh x+B'\sinh x = \frac{A'+B'}{2}e^{x}+\frac{A'-B'}{2}e^{-x}$$
Let $A=\frac{A'+B'}{2},\quad B=\frac{A'-B'}{2}$ and then we arrive back at $$y=Ae^x+Be^{-x}$$
Note that this works because all the constants used here are completely arbitrary.
Using the usual definitions of $\sinh x, \cosh x$, We can rewrite the second family of solutions, $$y(x) = A \sinh x + B \cosh x ,$$ to the given differential equation $y''(x) - y(x) = 0$ as $$y(x) = A \left(\frac{e^x - e^{-x}}{2}\right) + B \left(\frac{e^x + e^{-x}}{2}\right) = \left(\frac{A + B}{2}\right) e^x + \left(\frac{A - B}{2}\right) e^{-x} , $$ so every element solution in the second family can be written as an elements of the first family for appropriate choices of parameter. A similar computation shows that the reverse is true.
On the other hand, $A \cosh x + B \cosh x = (A + B) \cosh x$, so not every solution in the first and second families can be written in this form. (For one, $x \mapsto (A + B) \cosh x$ is even, but the solution $e^x$ is not.)
For a second-order ordinary differential equation (with reasonably nice coefficients) the solution set has $2$ real, independent parameters; in particular, there are infinitely many. If the equation is linear, that is, if it has the form $y''(x) + p(x) y'(x) + q(x) y(x) = r(x)$, then the solution space is affine. If it is homogeneous (that is, if it is affine with $r(x) = 0$), as is the case here, the solution space is a vector space, so as in this example, if we can find two linearly independent solutions $y_1(x), y_2(x)$, then the general solution is $$y(x) = C_1 y_1(x) + C_2 y_2(x), \qquad C_1, C_2 \in \Bbb R .$$ All of this generalizes immediately to higher-order equations.